Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $k = \dfrac{z + 1}{z^2 + 9z + 8} \times \dfrac{5z^2 + 20z - 300}{-3z - 30} $
Solution: First factor out any common factors. $k = \dfrac{z + 1}{z^2 + 9z + 8} \times \dfrac{5(z^2 + 4z - 60)}{-3(z + 10)} $ Then factor the quadratic expressions. $k = \dfrac {z + 1} {(z + 1)(z + 8)} \times \dfrac {5(z + 10)(z - 6)} {-3(z + 10)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {(z + 1) \times 5(z + 10)(z - 6) } { (z + 1)(z + 8) \times -3(z + 10)} $ $k = \dfrac {5(z + 10)(z - 6)(z + 1)} {-3(z + 1)(z + 8)(z + 10)} $ Notice that $(z + 1)$ and $(z + 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {5(z + 10)(z - 6)\cancel{(z + 1)}} {-3\cancel{(z + 1)}(z + 8)(z + 10)} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $k = \dfrac {5\cancel{(z + 10)}(z - 6)\cancel{(z + 1)}} {-3\cancel{(z + 1)}(z + 8)\cancel{(z + 10)}} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $k = \dfrac {5(z - 6)} {-3(z + 8)} $ $ k = \dfrac{-5(z - 6)}{3(z + 8)}; z \neq -1; z \neq -10 $